a) $FD//AC⇒\widehat{BDF}=\widehat{BCA}$ (đồng vị)
$DE//AB⇒\widehat{EDC}=\widehat{ABC}$ (đồng vị)
Ta có:
$\widehat{FDE}=180^o-\widehat{BDF}-\widehat{EDC}$
$\widehat{BAC}=180^o-\widehat{BCA}-\widehat{ABC}$
$⇒\widehat{FDE}=\widehat{BAC}$
b) $\widehat{BDF}=\widehat{BCA}$
$\widehat{EDC}=\widehat{ABC}$
$\widehat{FDE}=\widehat{BAC}$
mà $\widehat{BDF}+\widehat{EDC}+\widehat{FDE}=180^o$
⇒ $\widehat{ABC}+\widehat{BAC}+\widehat{BCA}=180^o$